杭电 Problem1865 1string

1sting

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4983    Accepted Submission(s): 1847

Problem Description

You will be given a string which only contains ‘1’; You can merge two adjacent ‘1’ to be ‘2’, or leave the ‘1’ there. Surly, you may get many different results. For example, given 1111 , you can get 1111, 121, 112,211,22. Now, your work is to find the total number of result you can get.

 

Input

The first line is a number n refers to the number of test cases. Then n lines follows, each line has a string made up of ‘1’ . The maximum length of the sequence is 200.

 

Output

The output contain n lines, each line output the number of result you can get .

 

Sample Input

3
1
11
11111

 

Sample Output

1
2
8

 

Author

z.jt

 

Source

2008杭电集训队选拔赛——热身赛

 

这题是斐波那契的运用，但是fib200会超掉储存，所以用大数的方式储存。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <iostream>
#define MAX_N   205
#define MAX(a, b) (a > b)? a: b
#define MIN(a, b) (a < b)? a: b
using namespace std;

int fib[MAX_N][MAX_N];
void init() {
    fib[1][0] = 1, fib[2][0] = 2;
    for (int i = 3; i < MAX_N; i++) {
        int q = 0, p = 0;
        for (int j = 0; j < MAX_N; j++) {
            q = fib[i - 1][j] + fib[i - 2][j] + p;
            fib[i][j] = q%10;
            p = q / 10;
        }
    }
}
int main() {
    int t;
    init();
    char s[MAX_N];
    scanf("%d", &t);
    while (t--) {
        scanf("%s", &s);
        int len = strlen(s);
        int i;
        for (i = 200; i >= 0 ; i--) {
            if (fib[len][i] != 0) {
                break;
            }
        }
        for (int j = i; j >= 0 ; j--) {
            printf("%d", fib[len][j]);
        }
        printf("\n");
    }
    return 0;
}